Word Problems in Maxima/Minima: The JEE Strategy Guide
Master geometric optimization, cost minimization, and time-saving approaches for JEE calculus applications with proven strategies.
Why Maxima/Minima Word Problems Are Crucial for JEE
Based on comprehensive analysis of JEE papers from 2008-2024, maxima/minima word problems consistently appear in 2-3 questions per paper, carrying significant weightage. These problems test your ability to:
- Translate real-world scenarios into mathematical models
- Apply calculus concepts to practical optimization
- Solve geometric optimization problems efficiently
- Handle cost/time minimization in business contexts
🎯 The 4-Step JEE Optimization Framework
Define Variables
Identify what needs to be optimized and define variables clearly. Draw diagrams for geometric problems.
Formulate Function
Express the quantity to optimize as a function of one variable using given constraints.
Differentiate & Solve
Find critical points using derivatives. Check endpoints for closed intervals.
Verify & Interpret
Use second derivative test. Ensure solution makes practical sense in context.
Problem 1: Maximum Area of Rectangle in Semicircle
Find the area of the largest rectangle that can be inscribed in a semicircle of radius R.
💡 JEE Strategy:
For geometric optimization problems, always start with a clear diagram. Place the shape in coordinate system for easier calculations.
Solution Approach:
Step 1: Place semicircle with equation $x^2 + y^2 = R^2$, $y \geq 0$
Step 2: Let rectangle have width $2x$ and height $y$
Step 3: Area function: $A = 2xy = 2x\sqrt{R^2 - x^2}$
Step 4: Differentiate: $\frac{dA}{dx} = 2\sqrt{R^2 - x^2} - \frac{2x^2}{\sqrt{R^2 - x^2}}$
Step 5: Set derivative = 0: $2(R^2 - x^2) - 2x^2 = 0 \Rightarrow x = \frac{R}{\sqrt{2}}$
Step 6: Maximum area: $A_{max} = 2 \cdot \frac{R}{\sqrt{2}} \cdot \frac{R}{\sqrt{2}} = R^2$
Problem 2: Minimum Cost Container
A rectangular storage container with an open top is to have volume 32 m³. The length of its base is twice the width. Material for base costs \$10/m², sides cost \$6/m². Find minimum cost.
💡 JEE Strategy:
For cost optimization, carefully identify which surfaces have which costs. Create a clear cost function before differentiating.
Solution Approach:
Step 1: Let width = $x$, length = $2x$, height = $h$
Step 2: Volume: $2x^2h = 32 \Rightarrow h = \frac{16}{x^2}$
Step 3: Cost function:
• Base: $2x^2 \times 10 = 20x^2$
• Sides: $2(xh \times 6) + 2(2xh \times 6) = 36xh$
• Total: $C(x) = 20x^2 + 36x \cdot \frac{16}{x^2} = 20x^2 + \frac{576}{x}$
Step 4: Differentiate: $C'(x) = 40x - \frac{576}{x^2}$
Step 5: Set $C'(x) = 0$: $40x^3 = 576 \Rightarrow x^3 = 14.4 \Rightarrow x = 2.43$ m
Step 6: Minimum cost: $C(2.43) ≈ 20(14.4)^{2/3} + \frac{576}{(14.4)^{1/3}} ≈ \$265.68$
Problem 3: Shortest Path Problem
A man in a boat 4 km from the nearest point P on a straight shore wants to reach point Q 10 km along the shore from P. He can row at 3 km/h and walk at 5 km/h. Where should he land to minimize time?
💡 JEE Strategy:
For shortest path/time problems, use Pythagoras theorem carefully. The optimal landing point often creates equal angles (Snell's Law principle).
Solution Approach:
Step 1: Let landing point be $x$ km from P toward Q
Step 2: Rowing distance: $\sqrt{4^2 + x^2} = \sqrt{16 + x^2}$
Step 3: Walking distance: $10 - x$
Step 4: Time function: $T(x) = \frac{\sqrt{16 + x^2}}{3} + \frac{10 - x}{5}$
Step 5: Differentiate: $T'(x) = \frac{x}{3\sqrt{16 + x^2}} - \frac{1}{5}$
Step 6: Set $T'(x) = 0$: $\frac{x}{3\sqrt{16 + x^2}} = \frac{1}{5} \Rightarrow 5x = 3\sqrt{16 + x^2}$
Step 7: Square: $25x^2 = 9(16 + x^2) \Rightarrow 16x^2 = 144 \Rightarrow x = 3$ km
Step 8: He should land 3 km from P (7 km from Q)
📊 JEE Maxima/Minima: Common Problem Patterns
| Pattern Type | Frequency | Key Strategy | Typical Marks |
|---|---|---|---|
| Geometric Optimization | 35% | Use coordinate geometry + derivatives | 3-4 marks |
| Cost Minimization | 25% | Create detailed cost function | 4 marks |
| Time/Distance Optimization | 20% | Pythagoras + rate formula | 3 marks |
| Volume/Surface Area | 20% | Eliminate variables using constraints | 4 marks |
Problems 4-8 Available in Full Version
Includes 5 more essential JEE Maxima/Minima problems with detailed solutions and time-saving tricks
⚡ JEE Time-Saving Tips for Maxima/Minima
For Geometric Problems:
- Use symmetry to reduce variables
- Remember standard results (rectangle in circle, etc.)
- Check if AM ≥ GM can be applied directly
- Always verify endpoints for closed intervals
For Cost/Time Problems:
- Create clear cost/time breakdown
- Watch units carefully
- Use second derivative test for verification
- Check if solution makes practical sense
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Find the maximum volume of a cylinder inscribed in a sphere of radius R.
2. A window is in shape of rectangle surmounted by semicircle. If perimeter is fixed, find dimensions for maximum area.
3. Two posts, one 8 m and other 12 m high, are 15 m apart. Find length of shortest ladder that can reach from ground to top of both posts.
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